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Work done by a constant force
When the point at which a force acts moves, the force is said to have done work.
When the force is constant, the work done is defined as the product of the force and
distance moved.
Consider the example in Figure 3.1, a force F acting at the angle  moves a body from
point A to point B.
Figure 3.1: Notation for work done by a force
The distance moved in the direction of the force is given by
So the work done by the force F is
Equation.1
If the body moves in the same direction as the force the angle is 0.0 so
Work done = Fs
When the angle is 90 then the work done is zero.
The SI units for work are Joules J (with force, F, in Newton's N and distance, s, in
metres m).
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Worked Example 3.1
How much work is done when a force of 5 kN moves its point of application 600mm
in the direction of the force.
Solution
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Worked Example 3.2
Find the work done in raising 100 kg of water through a vertical distance of 3m.
Solution
The force is the weight of the water, so
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Work done by a variable force
Forces in practice will often vary. In these cases Equation 3.1 cannot be used.
Consider the case where the force varies as in Figure 3.2
For the thin strip with width ds - shown shaded in Figure 3.2 - the force can be
considered constant at F. The work done over the distance ds is then
This is the area of the shaded strip.
The total work done for distance s is the sum of the areas of all such strips. This is the
same as the area under the Force-distance curve.
Figure 3.2: Work done by a variable force
So for a variable force
Equation.2
Clearly this also works for a constant force - the curve is then a horizontal line.
In general you must uses some special integration technique to obtain the area under a
curve. Three common techniques are the trapezoidal, mid-ordinate and Simpson's
rule. They are not detailed here but may be found in many mathematical text book.
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Energy
A body which has the capacity to do work is said to possess energy.
For example , water in a reservoir is said to possesses energy as it could be used to
drive a turbine lower down the valley. There are many forms of energy e.g. electrical,
chemical heat, nuclear, mechanical etc.
The SI units are the same as those for work, Joules J.
In this module only purely mechanical energy will be considered. This may be of two
kinds, potential and kinetic.
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Potential Energy
There are different forms of potential energy two examples are: i) a pile driver raised
ready to fall on to its target possesses gravitational potential energy while (ii) a coiled
spring which is compressed possesses an internal potential energy.
Only gravitational potential energy will be considered here. It may be described as
energy due to position relative to a standard position (normally chosen to be he earth's
surface.)
The potential energy of a body may be defined as the amount of work it would do if it
were to move from the its current position to the standard position.
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Formulae for gravitational potential energy
A body is at rest on the earth's surface. It is then raised a vertical distance h above the
surface. The work required to do this is the force required times the distance h.
Since the force required is it's weight, and weight, W = mg, then the work required is
mgh.
The body now possesses this amount of energy - stored as potential energy - it has the
capacity to do this amount of work, and would do so if allowed to fall to earth.
Potential energy is thus given by:
Equation 3
where h is the height above the earth's surface.
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Worked example 3.3
What is the potential energy of a 10kg mass:
a. 100m above the surface of the earth
b. at the bottom of a vertical mine shaft 1000m deep.
Solution
a)
b)
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Kinetic energy
Kinetic energy may be described as energy due to motion.
The kinetic energy of a body may be defined as the amount of work it can do before
being brought to rest.
For example when a hammer is used to knock in a nail, work is done on the nail by
the hammer and hence the hammer must have possessed energy.
Only linear motion will be considered here.
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Formulae for kinetic energy
Let a body of mass m moving with speed v be brought to rest with uniform
deceleration by a constant force F over a distance s.
Using Equation
And work done is given by
The force is F = ma so
Thus the kinetic energy is given by
Equation.4
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Kinetic energy and work done
When a body with mass m has its speed increased from u to v in a distance s by a
constant force F which produces an acceleration a, then from Equation 1. we know
multiplying this by m give an expression of the increase in kinetic energy (the
difference in kinetic energy at the end and the start)
Thus since F = ma
but also we know
So the relationship between kinetic energy can be summed up as
Work done by forces acting on a body = change of kinetic energy in the body
Equation 5
This is sometimes known as the work-energy theorem.
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Worked example 3.4
A car of mass 1000 kg traveling at 30m/s has its speed reduced to 10m/s by a constant
breaking force over a distance of 75m.
Find:
a. The cars initial kinetic energy
b. The final kinetic energy
c. The breaking force
Solution
a)
b)
c)
Change in kinetic energy = 400 kJ
By Equation 3.5 work done = change in kinetic energy so
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Conservation of energy
The principle of conservation of energy state that the total energy of a system remains
constant. Energy cannot be created or destroyed but may be converted from one form
to another.
Take the case of a crate on a slope. Initially it is at rest, all its energy is potential
energy. As it accelerates, some of it potential energy is converted into kinetic energy
and some used to overcome friction. This energy used to overcome friction is not lost
but converted into heat. At the bottom of the slope the energy will be purely kinetic
(assuming the datum for potential energy is the bottom of the slope.)
If we consider a body falling freely in air, neglecting air resistance, then mechanical
energy is conserved, as potential energy is lost and equal amount of kinetic energy is
gained as speed increases.
If the motion involves friction or collisions then the principle of conservation of
energy is true, but conservation of mechanical energy is not applicable as some
energy is converted to heat and perhaps sound.
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Worked Example 3.5
A cyclist and his bicycle has a mass of 80 kg. After 100m he reaches the top of a hill,
with slope 1 in 20 measured along the slope, at a speed of 2 m/s. He then free wheels
the 100m to the bottom of the hill where his speed has increased to 9m/s.
How much energy has he lost on the hill?
Solution
Figure 3.3: Dimensions of the hill in worked example 3.5
If the hill is 100m long then the height is:
So potential energy lost is
Increase in kinetic energy is
By the principle of conservation of energy
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Power
Power is the rate at which work is done, or the rate at which energy is used
transferred.
Equation.6
The SI unit for power is the watt W.
A power of 1W means that work is being done at the rate of 1J/s.
Larger units for power are the kilowatt kW (1kW = 1000 W = 103 W) and
the megawatt MW (1 MW = 1000000 W = 106 W).
If work is being done by a machine moving at speed v against a constant force, or
resistance, F, then since work doe is force times distance, work done per second is Fv,
which is the same as power.
Equation 7
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Worked Example 3.6
A constant force of 2kN pulls a crate along a level floor a distance of 10 m in 50s.
What is the power used?
Solution
Alternatively we could have calculated the speed first
and then calculated power
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Worked Example 3.7
A hoist operated by an electric motor has a mass of 500 kg. It raises a load of 300 kg
vertically at a steady speed of 0.2 m/s. Frictional resistance can be taken to be
constant at 1200 N.
What is the power required?
Solution
From Equation 3.7
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Worked example 3.8
A car of mass 900 kg has an engine with power output of 42 kW. It can achieve a
maximum speed of 120 km/h along the level.
a. What is the resistance to motion?
b. If the maximum power and the resistance remained the same what would be
the maximum speed the car could achieve up an incline of 1 in 40 along the
slope?
Solution
Figure 3.4: Forces on the car on a slope in Worked Example 3.8
First get the information into the correct units:
a) Calculate the resistance
b)
Or in km/h
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Moment, couple and torque
The moment of a force F about a point is its turning effect about the point.
It is quantified as the product of the force and the perpendicular distance from the
point to the line of action of the force.
Figure 3.4: Moment of a force
In Figure 3.5 the moment of F about point O is
Equation 8
A couple is a pair of equal and parallel but opposite forces as shown in Figure 3.6:
Figure 3.6: A couple
The moment of a couple about any point in its plane is the product of one force and
the perpendicular distance between them:
Equation.9
Example of a couple include turning on/off a tap, or winding a clock.
The SI units for a moment or a couple are Newton metres, Nm.
In engineering the moment of a force or couple is know as torque. A spanner
tightening a nut is said to exert a torque on the nut, similarly a belt turning a pulley
exerts a torque on the pulley.
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Work done by a constant torque
Let a force F turn a light rod OA with length r through an angle of  to position OB,
as shown in Figure 3.7.
Figure 3.7: Work done by a constant torque
The torque TQ exerted about O is force times perpendicular distance from O.
Equation 10
Now work done by F is
s is the arc of the circle, when is measure in radians
Equation 11
The work done by a constant torque TQ is thus the product of the torque and the angle
through which it turns (where the angle is measured in radians.)
As the SI units for work is Joules, TQ must be in Nm
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Power transmitted by a constant torque
Power is rate of doing work. It the rod in Figure 3.7 rotates at n revolutions per
second, then in one second the angle turned through is
radians, and the work done per second will be, by Equation 11
as angular speed is
then
Equation 12
The units of power are Watts, W, with n in rev/s,  in rad/s and TQ in Nm.
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Worked Example 3.9
A spanner that is used to tighten a nut is 300mm long. The force exerted on the end of
a spanner is 100 N.
a. What is the torque exerted on the nut?
b. What is the work done when the nut turns through 30 ?
Solution
a)
Calculate the torque by Equation 10
b)
Calculate the work done by Equation 3.11
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Worked Example 3.10
An electric motor is rated at 400 W. If its efficiency is 80%, find the maximum torque
which it can exert when running at 2850 rev/min.
Solution
Calculate the speed in rev/s using Equation 12
Calculate the power as the motor is 80% efficient
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Work done by a variable torque
In practice the torque is often variable. In this case the work done cannot be calculated
by Equation 3.11, but must be found in a similar way to that used for a variable force
(see earlier.)
Figure 3.8: Work done by a variable torque
The work done when angular displacement is dis TQd. This is the area of the shaded
strip in Figure 3.8. the total work done for the angular displacement  is thus the area
under the torque/displacement graph.
For variable torque
Equation 13
As with variable forces, in general you must uses some special integration technique
to obtain the area under a curve. Three common techniques are the trapezoidal, midordinate and Simpson's rule. They are not detailed here but may be found in many
mathematical text book.
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Worked Example 3.11
A machine requires a variable torque as shown in Figure 3.9, Find:
a. The work done per revolution
b. The average torque over one revolution
c. The power required if the machine operates at 30 rev/min
Figure 3.9: Torque requirement for Worked Example 3.12
Solution
a)
From Equation 13
for one revolution
b)
Average torque is the average height of figure OABCDE = area /2
c)